package cn.hycat.leetcode.middle;

/**
 * @author 吕泽浩业
 * @version 1.0
 */
/*
200. 岛屿数量
        给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
        岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
        此外，你可以假设该网格的四条边均被水包围。

        示例 1：
        输入：grid = [
        ["1","1","1","1","0"],
        ["1","1","0","1","0"],
        ["1","1","0","0","0"],
        ["0","0","0","0","0"]
        ]
        输出：1

        输入：grid = [
        ["1","1","0","0","0"],
        ["1","1","0","0","0"],
        ["0","0","1","0","0"],
        ["0","0","0","1","1"]
        ]
        输出：3

        m == grid.length
        n == grid[i].length
        1 <= m, n <= 300
        grid[i][j] 的值为 '0' 或 '1'
*/
public class L200岛屿数量 {
    public static void main(String[] args) {
        char[][] grid = {
                {'1','1','0','0','0'},
                {'1','1','0','0','0'},
                {'0','0','1','0','0'},
                {'0','0','0','1','1'}};
        System.out.println(numIslands(grid));

        print(grid);
    }
    private static int ans = 0;
    public static int numIslands(char[][] grid) {
        int N = grid.length;
        int M = grid[0].length;
        for(int i = 0; i < N; i++) {
            for(int j  = 0; j < M; j++) {
                if(grid[i][j] == '1') {
                    dfs(grid, i, j);
                    ans++;
                }
            }
        }
        return ans;
    }

    public static void dfs(char[][] grid, int i, int j) {
        if(i > grid.length - 1 || i < 0 || j > grid[i].length - 1 || j < 0
                || grid[i][j] != '1') return;
        grid[i][j] = '0';
        dfs(grid, i+1, j);
        dfs(grid, i, j+1);
        dfs(grid, i-1, j);
        dfs(grid, i, j-1);
    }

    public static void print(char[][] grid) {
        int N = grid.length;
        int M = grid[0].length;
        for(int i = 0; i < N; i++) {
            for(int j  = 0; j < M; j++) {
                System.out.print(grid[i][j] + " ");
            }
            System.out.println();
        }
    }
}
